Strand 2 — Geometry & Trigonometry

Co-ordinate Geometry of the Line

5th Year · 6th Year (Leaving Cert)

✓Recall and apply different forms of the equation of a line, including general, slope-intercept, and point-slope forms.
✓Calculate the perpendicular distance from a given point to a given line (Higher Level).
✓Determine the acute angle between two given lines (Higher Level).
✓Calculate the area of a triangle given the co-ordinates of its vertices.

Key concepts

Equation Forms of a Line

The equation of a line can be expressed in several forms. The slope 'm' of a line passing through two points (x1, y1) and (x2, y2) is given by m = (y2 - y1) / (x2 - x1).

m = (y2 - y1) / (x2 - x1)

Point-Slope Form

This form is used when you know the slope 'm' of a line and a point (x1, y1) it passes through.

y - y1 = m(x - x1)

Slope-Intercept Form

This form is useful when the slope 'm' and the y-intercept 'c' (the point where the line crosses the y-axis) are known.

y = mx + c

General Form

The general form of the equation of a line is ax + by + c = 0, where a, b, and c are integers, and 'a' is usually positive. This form is often required for calculations involving distance from a point to a line.

ax + by + c = 0

Parallel and Perpendicular Lines

Two lines are parallel if they have the same slope (m1 = m2). Two lines are perpendicular if the product of their slopes is -1 (m1 * m2 = -1), provided neither line is vertical.

Parallel: m1 = m2; Perpendicular: m1 * m2 = -1

Distance from a Point to a Line (Higher Level)

The perpendicular distance 'd' from a point (x1, y1) to a line given in the general form ax + by + c = 0 is calculated using the formula. Ensure the line equation is in general form before applying.

d = |ax1 + by1 + c| / sqrt(a^2 + b^2)

Angle Between Two Lines (Higher Level)

If two lines L1 and L2 have slopes m1 and m2 respectively, the acute angle 'theta' between them can be found using the formula. The absolute value ensures the acute angle is found. If 1 + m1*m2 = 0, the lines are perpendicular (theta = 90 degrees).

tan(theta) = |(m1 - m2) / (1 + m1*m2)|

Area of a Triangle

To find the area of a triangle with vertices A(x1, y1), B(x2, y2), and C(x3, y3), a common method is to translate one vertex (e.g., A) to the origin (0,0) by subtracting its coordinates from all vertices. Let the new coordinates of B and C be (X1, Y1) and (X2, Y2) respectively. The area is then given by the formula.

Area = 1/2 |X1Y2 - X2Y1|

Key facts to remember

1The slope 'm' of a line passing through (x1, y1) and (x2, y2) is (y2 - y1) / (x2 - x1).
2Parallel lines have equal slopes (m1 = m2).
3Perpendicular lines have slopes whose product is -1 (m1 * m2 = -1), provided neither line is vertical.
4The general form of a line is ax + by + c = 0.
5The distance from (x1, y1) to ax + by + c = 0 is |ax1 + by1 + c| / sqrt(a^2 + b^2) (HL).
6The acute angle 'theta' between two lines with slopes m1 and m2 is given by tan(theta) = |(m1 - m2) / (1 + m1*m2)| (HL).
7The area of a triangle with vertices (x1, y1), (x2, y2), (x3, y3) can be found by translating one vertex to the origin, then using 1/2 |X1Y2 - X2Y1| for the other two translated vertices (X1, Y1) and (X2, Y2).

Worked examples

Example 1

Find the equation of the line L that passes through the point (2, -3) and is perpendicular to the line 3x - 2y + 5 = 0. Give your answer in the form ax + by + c = 0.

I1. Find the slope of the given line 3x - 2y + 5 = 0.

II Rearrange to slope-intercept form: 2y = 3x + 5 => y = (3/2)x + 5/2.

III The slope of the given line, m1, is 3/2.

IV2. Determine the slope of the perpendicular line L.

V For perpendicular lines, m1 * m2 = -1.

VI (3/2) * m2 = -1 => m2 = -2/3.

VII3. Use the point-slope form y - y1 = m(x - x1) with the point (2, -3) and slope m = -2/3.

VIII y - (-3) = (-2/3)(x - 2)

9 y + 3 = (-2/3)(x - 2)

104. Clear the denominator and rearrange into the general form ax + by + c = 0.

11 3(y + 3) = -2(x - 2)

12 3y + 9 = -2x + 4

13 2x + 3y + 9 - 4 = 0

14 2x + 3y + 5 = 0.

Answer

2x + 3y + 5 = 0

Always ensure the final equation is in the requested form, typically ax + by + c = 0 with 'a' positive.

Example 2

Calculate the perpendicular distance from the point P(3, -1) to the line L: 4x - 3y + 7 = 0.

I1. Identify the coordinates of the point (x1, y1) and the coefficients a, b, c from the line equation.

II Point (x1, y1) = (3, -1).

III Line ax + by + c = 0 is 4x - 3y + 7 = 0, so a = 4, b = -3, c = 7.

IV2. Apply the distance formula: d = |ax1 + by1 + c| / sqrt(a^2 + b^2).

V d = |(4)(3) + (-3)(-1) + 7| / sqrt(4^2 + (-3)^2)

VI d = |12 + 3 + 7| / sqrt(16 + 9)

VII d = |22| / sqrt(25)

VIII d = 22 / 5.

Answer

22/5 units

The distance is always a positive value, hence the absolute value in the numerator.

Example 3

(a) Find the acute angle between the lines L1: y = 2x + 1 and L2: x + 3y - 6 = 0. (b) Calculate the area of the triangle with vertices A(1, 2), B(5, 4), and C(3, 8).

I(a) Steps for finding the angle:

II 1. Find the slopes m1 and m2 of the two lines.

III L1: y = 2x + 1 => m1 = 2.

IV L2: x + 3y - 6 = 0 => 3y = -x + 6 => y = (-1/3)x + 2 => m2 = -1/3.

V 2. Apply the formula tan(theta) = |(m1 - m2) / (1 + m1*m2)|.

VI tan(theta) = |(2 - (-1/3)) / (1 + (2)(-1/3))|

VII tan(theta) = |(2 + 1/3) / (1 - 2/3)|

VIII tan(theta) = |(7/3) / (1/3)|

9 tan(theta) = |7|

10 tan(theta) = 7.

11 3. Calculate theta.

12 theta = tan^-1(7) approx 81.87 degrees (to 2 decimal places).

13(b) Steps for finding the area of the triangle:

14 1. Translate one vertex to the origin. Let's translate A(1, 2) to (0, 0).

15 Subtract (1, 2) from all vertices:

16 A' = (1-1, 2-2) = (0, 0)

17 B' = (5-1, 4-2) = (4, 2)

18 C' = (3-1, 8-2) = (2, 6)

19 2. Use the formula Area = 1/2 |X1Y2 - X2Y1| with B'(X1, Y1) = (4, 2) and C'(X2, Y2) = (2, 6).

20 Area = 1/2 |(4)(6) - (2)(2)|

21 Area = 1/2 |24 - 4|

22 Area = 1/2 |20|

23 Area = 10.

Answer

(a) The acute angle is approximately 81.87 degrees. (b) The area of the triangle is 10 square units.

For the angle, ensure you find the acute angle by taking the absolute value. For the area, translating a vertex to the origin simplifies the calculation significantly.

Common mistakes

✗Not rearranging the line equation into the general form ax + by + c = 0 before using the distance formula.
✗Incorrectly calculating the negative reciprocal for perpendicular slopes (e.g., just changing the sign instead of flipping the fraction and changing the sign).
✗Forgetting the absolute value in the distance formula or the angle formula, leading to negative distances or angles.
✗Mixing up x and y coordinates when calculating slope or substituting into equations.
✗Arithmetic errors, especially with fractions or negative numbers, which are common in co-ordinate geometry.

Exam tips

★Always write down the formula you are using before substituting values; this helps with partial credit if there's a calculation error.
★Check your slopes: a positive slope should go 'uphill' from left to right, a negative slope 'downhill'. This can help catch sign errors.
★If a question involves perpendicular lines, immediately think m1 * m2 = -1. If parallel, m1 = m2.
★For area of a triangle, translating one vertex to the origin often simplifies calculations and reduces the chance of error compared to the full shoelace formula.
★Be careful with signs when substituting coordinates into formulas, especially when dealing with negative numbers.

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